many angleswith point on the arc of circumcircles

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many angleswith point on the arc of circumcircles

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#1 h Jul 25, 2018, 3:13 PM • 2 Y

Y by Adventure10, Mango247

Consider triangle $ABC$ such that $AB \le AC$ . Point $D$ on the arc $BC$ of thecircumcirle of $ABC$ not containing point $A$ and point $E$ on side $BC$ are such that $\angle BAD = \angle CAE < \frac12 \angle BAC$ . Let $S$ be the midpoint of segment $AD$ . If $\angle ADE = \angle ABC - \angle ACB$ prove that $\angle BSC = 2 \angle BAC$ .

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#2 h Jun 13, 2019, 10:25 AM • 1 Y

Y by Adventure10

Let $AD$ meets $BC$ in point $K$ . $\angle ABC = \angle ADC$ so by condition $\angle EDC = \angle ACB = \angle ADB$ so $DE$ and $DA$ are isogonal conjugate in triangle $BDC$ and $AE$ and $DA$ are isogonal conjugate in triangle $ABC$ . It is well-known that $(\frac{BD}{CD})^2 = \frac{BK}{CK} \frac{BE}{CE} = (\frac{BA}{CA})^2$ so $BD*AC = CD*AB$ and $ABCD$ is harmonic.
Now let tangents to $B$ and $C$ and $AD$ meets in point $F$ .
$\angle BFC = \angle AFC + \angle BFA = 180 - \angle DAC - \angle DCA - \angle DCF + 180 - \angle BAD - \angle ABD - \angle DBF = 360 - 2\angle BAC - 180 = 180 - 2\angle BAC$
and
$\angle FBS + \angle SCF = \angle ABD - \angle CBD + \angle BAD + \angle ACD - \angle BCD + \angle DAC = 180 - \angle BAC + \angle BAC = 180$
Here we use that BC is symedian in triangle $ABD$ and $CB$ is symedian in triangle $ACD$ .
So quadrilateral $BSCF$ is cyclic. And $\angle BSC = 180 - \angle BFC = 2\angle BAC$ .
Done.

This post has been edited 3 times. Last edited by Daniil02, Jun 13, 2019, 10:28 AM

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#3 h Jan 18, 2021, 3:39 PM • 3 Y

Y by Mango247, Mango247, Mango247

Another (easier) solution???

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#4 h Jan 18, 2021, 5:31 PM • 2 Y

Y by sttsmet, Mango247

Show $BD \cdot AC =CD \cdot AB (*)$ like @2above , we can apply Ptolemy's Theorem to get $BC\cdot AD = AB\cdot CD + AC \cdot BD$ . Combine with $(*)$ to get $\frac{SD}{CD}=\frac{AD}{2\cdot CD}=\frac{BA}{BC}$ . By cyclic quadrilateral, we have $\angle ABC= \angle SDC$ . Then, $\triangle ABC \sim \triangle SDC$ .
Let $O$ be the circumcenter. $\angle OBC=90^{ \circ} -\angle BAC=\angle OSC.$ Therefore, $BSOC$ is cyclic. Thus, $2 \angle BAC= \angle BOC= \angle BSC$ .
$\blacksquare$

This post has been edited 1 time. Last edited by agwwtl03, Jan 18, 2021, 5:33 PM
Reason: typo

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#5 h Jun 9, 2023, 7:58 PM • 1 Y

Y by egxa

Let $O$ be the circumcenter of triangle $ABC$ . And let $R \in AD, R\in (BOC)$ . We'll prove that $R=S$ $\iff$ $AR=RD$ $\iff$ $OR \perp AD$ .
If $DE \cap (ABC)=K$ and $AE \cap (ABC)$ , $AK \parallel BC \parallel DL$ .
$OE \perp BC$
$AD,OE,KL$ are concurrence at $N$ .
$OE\cap KA=Q, CT \cap KA=P$ , $\angle PCN=90=\angle PQN \implies P,Q,C,N$ are cyclic.
$\angle OBC=\angle BCQ=\angle CQA=\angle CNP= \angle CNO \implies N \in BOC$
$\implies B,R,O,C,N$ are cyclic.
$\angle BRN=\angle BAC \implies OR \perp AD$ .

This post has been edited 1 time. Last edited by bin_sherlo, Jun 9, 2023, 7:59 PM

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#6 h Jun 10, 2023, 9:11 AM

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After consider $E$ is Dumpty point from $C$ on $ADC$ the question is trivial

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